Hydrogen sulfide decomposes according to the following reaction, for which Kc = 9.30 10-8 at 700°C
2 H2S(g) 2 H2(g) + S2(g)
If 0.51 mol H2S is placed in a 3.0 L container, what is the equilibrium concentration of H2(g) at 700°C?
0.51 mol/3 L = (H2S) in mols/L = M
.........2H2S --> 2H2 + S2
I........0.17......0.....0
C.........-2x.....2x.....x
E......0.17-2x....2x.....x
Substitute the E line into the Kq expression and solve for x. Then evaluate each specie.
To find the equilibrium concentration of H2(g) at 700°C, we can use the equilibrium constant expression and the given information.
The equilibrium constant expression for the given reaction is:
Kc = ([H2]^2[S2])/[H2S]^2
Here, [H2] represents the equilibrium concentration of H2(g), [S2] represents the equilibrium concentration of S2(g), and [H2S] represents the initial concentration of H2S(g).
We are given the equilibrium constant (Kc) as 9.30 * 10^(-8) and the initial concentration of H2S(g) as 0.51 mol in a 3.0 L container.
Since the stoichiometric coefficient of H2S is 2 in the balanced equation, the initial concentration of H2S is halved to 0.51/2 = 0.255 mol/L.
Let's assume the equilibrium concentration of H2(g) is x mol/L.
Using the equilibrium constant expression, we can write:
9.30 * 10^(-8) = (x^2)([S2])/[H2S]^2
Since the concentration of S2(g) is not given, we can ignore it for now and focus on finding the equilibrium concentration of H2(g).
Rearranging the equation and substituting the values:
x^2 = (9.30 * 10^(-8)) * [H2S]^2
x^2 = (9.30 * 10^(-8)) * (0.255)^2
x^2 = 5.97755 * 10^(-10)
x ≈ 2.445 * 10^(-5) mol/L
Therefore, the equilibrium concentration of H2(g) at 700°C is approximately 2.445 * 10^(-5) mol/L.